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gavmur [86]
2 years ago
8

An action potential arriving at the presynaptic terminal causes what to occur? multiple choice voltage-gated calcium ion channel

s open, and calcium ions diffuse into the cell. voltage-gated sodium ion channels open, and sodium ions diffuse into the cell. acetylcholine diffuses into the cell. ligand-gated sodium channels open, and sodium ions diffuse out of the cell. voltage-gated sodium ion channels open, and sodium ions diffuse out of the cell.
Chemistry
1 answer:
Mazyrski [523]2 years ago
5 0
The answer to this question would be: calcium ions to diffuse into the cell.

The presynaptic neuron would need to release neurotransmitter to the postsynaptic neuron. To do this, the action potential will cause calcium ion diffuse into the cell causing the vesicle with the neurotransmitter to migrate to the cell membrane and release its content. 
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A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m
Phoenix [80]
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
Mr = 27.2
Mass of empirical formula = 14
Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
7 0
2 years ago
Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10
Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

<u><em>a. 0.10 M acetic acid/0.25 M sodium acetate </em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em>b. 0.25 M acetic acid/0.10 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

<u><em>c. 0.080 M acetic acid/0.20 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH =  4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em></em></u>

<u><em>d. 0.20 M acetic acid/0.080 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0
2 years ago
Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) form a self-igniting mixture that has been used as a rocket propellant. The rea
Alexxx [7]

Answer:

Explanation:

An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.

Chemical equation:

1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O

2) Hydrazine ( N₂H₄)  is being oxidized

Dinitrogen tetroxide N₂O₄ is being reduced

3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)

5 0
2 years ago
How many moles of hydrogen gas are produced when 0.066 mole of sodium is completely reacted?
SVEN [57.7K]
Base on my research the complete reaction of hydrogen and sodium are form by this equation  2Na + 2H2O = 2NaOH +H2 If we base from this equation it shows you already the ratio of the moles of the product and reactants taking part in the reaction

1 mole of Na r/w 1 mole of water to produce .5 mole of Hydrogen 

Therefore, .066 mole of Na produces .033 moles of H2
6 0
2 years ago
Read 2 more answers
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
2 years ago
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